In previous post about material balances, we can see that the flow of any component was in the same ratio to the flow of any other component, as the ratio of the concentrations of the two components. If one component passes unchanged through a process unit it can be used to tie the inlet and outlet compositions.

This method is very helpful for handling combustion calculations if the nitrogen in the combustion air is employed as the tie component and flows through unreacted.

Let’s see example below.

*Carbon dioxide is added at a rate of 20 kg/h to an air stream and the air is sampled at a*

*sufficient distance downstream to ensure complete mixing. If the analysis shows 0.5%v/v CO _{2}. Calculate the air-flow rate.*

__Solution__

Normal carbon dioxide content of air is 0.03%. So, the mixing is illustrated in figure below.

We use basis of kmol/h as the concentration is expressed in volume fraction.

CO_{2} introduced = mass flow rate / molecular weight = 20 / 44 = 0.45 kmol/h

Let *X *be the air flow.

**CO _{2} in = CO_{2} out**

CO2 in = 0.03% *X *+ 0.45 = 0.0003 *X *+ 0.45

CO2 out = 0.5% *X *= 0.005 *X*

0.0003 *X *+ 0.45 = 0.005 *X*

*X* = 95.74 kmol/h

*X* = 95.74 x 29 = __2776 kg/h__

I hope you find this simple post is useful.

For your information, I made a series of material balances calculation post. Please check also my other posts: